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Maths-

General

Easy

Question

The transformed equation of $7 x to the power of 2 end exponent plus 2 square root of 3 x y plus 9 y to the power of 2 end exponent equals 8$ when the axes are rotated through an angle $fraction numerator text π end text over denominator text 3 end text end fraction$ is

$3 X^{2} - Y^{2} + 2 \sqrt{2} Y - 6 = 0$
$5 X^{2} + 3 Y^{2} = 5$
$5 X^{2} + 3 Y^{2} = 4$
$4 X^{2} + 3 Y^{2} = 6$

The correct answer is: $5 X to the power of 2 end exponent plus 3 Y to the power of 2 end exponent equals 4$

x = X cos θ -Ysin θ, y = X sin θ +Ycos θ
$x equals blank fraction numerator X minus square root of 3 Y over denominator 2 end fraction comma blank y equals fraction numerator square root of 3 X plus Y over denominator 2 end fraction$
$7 open parentheses fraction numerator X minus square root of 3 Y over denominator 2 end fraction close parentheses to the power of 2 end exponent plus 2 open parentheses fraction numerator X minus square root of 3 Y over denominator 2 end fraction close parentheses open parentheses fraction numerator square root of 3 X plus Y over denominator 2 end fraction close parentheses plus 9 open parentheses fraction numerator square root of 3 X plus Y over denominator 2 end fraction close parentheses to the power of 2 end exponent equals$ 8
$5 X to the power of 2 end exponent plus 3 Y to the power of 2 end exponent equals 4$

Related Questions to study

The transformed equation of $x to the power of 2 end exponent minus y to the power of 2 end exponent plus 2 x plus 4 y equals 0$ when the origin is shifted to the point (-1,2) is

The transformed equation of $x to the power of 2 end exponent minus y to the power of 2 end exponent plus 2 x plus 4 y equals 0$ when the origin is shifted to the point (-1,2) is

The point to which the origin should be shifted in order to eliminate x and y terms in the equation $4 x to the power of 2 end exponent plus 9 y to the power of 2 end exponent minus 8 x plus 36 y plus 4$ = 0

The point to which the origin should be shifted in order to eliminate x and y terms in the equation $4 x to the power of 2 end exponent plus 9 y to the power of 2 end exponent minus 8 x plus 36 y plus 4$ = 0

Orthocentre of the triangle is (2,1) and circumcentre is $open parentheses fraction numerator text 7 end text over denominator text 2 end text end fraction text , end text fraction numerator text 5 end text over denominator text 2 end text end fraction close parentheses$ , then its nine point circle centre is

Orthocentre of the triangle is (2,1) and circumcentre is $open parentheses fraction numerator text 7 end text over denominator text 2 end text end fraction text , end text fraction numerator text 5 end text over denominator text 2 end text end fraction close parentheses$ , then its nine point circle centre is

parallel

If a vertex of a $capital delta to the power of l e end exponent$ is (1,1) and the midpoints of two sides through this vertex are (-1,2)(3,2) then centroid of the triangle is

If a vertex of a $capital delta to the power of l e end exponent$ is (1,1) and the midpoints of two sides through this vertex are (-1,2)(3,2) then centroid of the triangle is

$I f blank A equals open parentheses 1 comma negative 1 close parentheses comma B equals open parentheses 2 comma 1 close parentheses comma C equals open parentheses 3 comma 0 close parentheses comma blank t h e n blank t h e blank c i r c u m minus r a d i u s blank o f blank increment A B C blank i s$

$I f blank A equals open parentheses 1 comma negative 1 close parentheses comma B equals open parentheses 2 comma 1 close parentheses comma C equals open parentheses 3 comma 0 close parentheses comma blank t h e n blank t h e blank c i r c u m minus r a d i u s blank o f blank increment A B C blank i s$

$T h e blank p o i n t s blank open parentheses a comma 0 close parentheses open parentheses 0 comma b close parentheses open parentheses 1 comma 1 close parentheses blank a r e blank c o l l i n e a r blank i f blank$

$T h e blank p o i n t s blank open parentheses a comma 0 close parentheses open parentheses 0 comma b close parentheses open parentheses 1 comma 1 close parentheses blank a r e blank c o l l i n e a r blank i f blank$

parallel

$I f fraction numerator pi over denominator 2 end fraction less than alpha less than pi comma blank t h e n blank t h e blank d i s t a n c e blank b e t w e e n blank t h e blank p o i n t s blank open parentheses t a n blank alpha comma 2 close parentheses comma$ $open parentheses 0 comma 1 close parentheses blank i s$

$I f fraction numerator pi over denominator 2 end fraction less than alpha less than pi comma blank t h e n blank t h e blank d i s t a n c e blank b e t w e e n blank t h e blank p o i n t s blank open parentheses t a n blank alpha comma 2 close parentheses comma$ $open parentheses 0 comma 1 close parentheses blank i s$

From the figure describing photoelectric effect we may infer correctly that

From the figure describing photoelectric effect we may infer correctly that

Physics-General

In Thomson spectrograph experiment, four positive ions P,Q,R and S are situated on Y-X curve a shown in the figure

In Thomson spectrograph experiment, four positive ions P,Q,R and S are situated on Y-X curve a shown in the figure

Physics-General

parallel

In a parabola spectrograph, the velocities of four positive ions P,Q,R and S are v₁, v₂, v₃ and v₄ respectively

In a parabola spectrograph, the velocities of four positive ions P,Q,R and S are v₁, v₂, v₃ and v₄ respectively

Physics-General

The maximum value of stopping potential in the following diagram is

The maximum value of stopping potential in the following diagram is

Physics-General

In the diagram a graph between the intensity of X-rays emitted by a molybdenum target and the wavelength is shown, when electrons of 30 keV are incident on the target. In the graph one peak is of $K subscript alpha end subscript$ line and the other peak is of $K subscript beta end subscript$ line

In the diagram a graph between the intensity of X-rays emitted by a molybdenum target and the wavelength is shown, when electrons of 30 keV are incident on the target. In the graph one peak is of $K subscript alpha end subscript$ line and the other peak is of $K subscript beta end subscript$ line

Physics-General

parallel

The graph between the square root of the frequency of a specific line of characteristic spectrum of X-rays and the atomic number of the target will be

The graph between the square root of the frequency of a specific line of characteristic spectrum of X-rays and the atomic number of the target will be

Physics-General

In the graph given below. If the slope is $4.12 cross times 1 0 to the power of negative 15 end exponent$ V-sec, then value of ‘h’ should be

In the graph given below. If the slope is $4.12 cross times 1 0 to the power of negative 15 end exponent$ V-sec, then value of ‘h’ should be

Physics-General

The variation of wavelength $lambda$ of the $K subscript alpha end subscript$ line with atomic number Z of the target is shown by the following curve of

The variation of wavelength $lambda$ of the $K subscript alpha end subscript$ line with atomic number Z of the target is shown by the following curve of

Physics-General

parallel

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